(a1^2+a2^2+……+an^2)(b1^2+b2^2+……+bn^2)≥(a1b1+a2b2+……+anbn)^2证：考虑这个代数式：(a1t-b1)^2+(a2t-b2)^2+……+(ant-bn)^2显然有(a1t-b1)^2+(a2t-b2)^2+……+(ant-bn)^2≥0左边拆开,(a1^2+a2^2+……+an^2)...